# Is $f(x) = |\arctan(x)|$ a norm on $\mathbb{R}$?

Is $f(x) = |\arctan(x)|$ a norm on $\mathbb{R}$?

Im checking if the properties of a norm holds for $f(x) = |\arctan(x)|$.

$1. \ f(x) \ge 0 \Leftrightarrow |\arctan(x)| \ge 0 \\ 2. \ f(x)=0 \Leftrightarrow |\arctan(x)| =0 \Leftrightarrow x=0 \\$

But does $f(\lambda x)=|\arctan(\lambda x)|\Leftrightarrow |\lambda||\arctan(x)|?$ For some $\lambda \in \mathbb{K}$.

Also, how would I check if $|\arctan(x+y)| \le |\arctan(x)|+|\arctan(y)|$?

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Have you tried checking whether $|\arctan(\lambda x)|=|\lambda||\arctan(x)|$ for some actual values of $\lambda$ and $x$? – Chris Eagle May 17 '12 at 15:30
Echoing Chris, take an example where $\lambda\neq 0$ and $x\neq 0$. – Jonas Meyer May 17 '12 at 15:31
You also need $f(x)=0 \implies x=0$, don't you? It is true, but you should state it. – Ross Millikan May 17 '12 at 15:33
Thanks, so by giving (1) counterexample with actual values of $\lambda$ and $x$ and showing the equality does not hold, is sufficient? Thanks Ross Millikan, I stated it above :) – Steven May 17 '12 at 15:48

$\arctan$ can be used a distance, but not as a norm. (As in $d(x,y) = |\arctan(x)-\arctan(y)|$, which produces an incomplete metric space.)

$\arctan$ is bounded, so it cannot satisfy $|\arctan(\lambda x)| = |\lambda||\arctan(x)|$.

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Thanks copper.hat, another way to see it compared to substituting in values! – Steven May 17 '12 at 15:51

arctan is an odd function and its values at positive arguments are positive, so questions about the inequality $|\arctan(x+y)|\le|\arctan x| + |\arctan y|$ are reducible to questions where $x$ and $y$ are positive and no absolute values are considered.

So if $x,y>0$, how do we know $\arctan (x+y) \le \arctan x + \arctan y$?

Just notice that the growth rate of the arctan function gets smaller as $x$ gets bigger. If we fix $x>0$ and let $y$ grow from $0$ to some positive number, the right side of the inequality is always growing faster than the left side, since on the right side you're taking the arctan of something closer to $0$.

Therefore the inequality is true.

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Thanks, nice way to put it! – Steven May 17 '12 at 15:52